# Rational numbers and leibniz law

Leibniz law says $a = b \implies f(a) = f(b)$. Unfortunately this law seems to fail for rational numbers e.g. ${1 \over 2} = {2 \over 4}$ but $numerator({1 \over 2}) \neq numerator({2 \over 4})$. I know that you can say $1 \over 2$ is just representation of "true" rational number and equality we use is just equivalence relation not real equality but this representation is how we think about rational numbers and how we define them in abstract algebra.

Question is: Is there some logically precise (e.g. first order logic) and "true" definition of natural number that respects Leibniz law and makes ${1 \over 2}$ and ${2 \over 4}$ truly equal not just equivalent.

Edit:

To rephrase the question: how do I define rational numbers to avoid problems with representation. In other words is there some axiomatic definition of rationals. Up to this point I've only seen algebraic definitions with pairs and equivalence classes.

You are confusing the object with its various names.

As an object, $\frac12=\frac24$. But you can present it differently. More specifically, in order for the numerator function to be well-defined, you need to choose a representation for each rational first.

Similarly, $1+1=2$, but the length of these two expressions is different. Names are syntax, objects are semantics. Leibniz's law is about semantics, not syntax.

• I do not agree that $1 \over 2$ is syntax. Syntax is described by axioms and $1 \over 2$ is model of those axioms so it is semantics. Question that is still open is how do I define rational number to avoid problems with representation. In other words is there some axiomatic definition of rationals. Up to this poin I've only seen agebraic definitions with pairs and equivalence classes. – Trismegistos Oct 9 '15 at 12:02
• Yes, you use equivalence classes. Then $\frac12$ should be seen as the appropriate equivalence class, rather than the actual ordered pair creating it. – Asaf Karagila Oct 9 '15 at 12:20

Your problem is that you are separating the number $\frac12$ into a pair of numbers $(1,2)$. Your law would then still be true, since $(1,2)\neq(2,4)$ and there would be no reason to suppose that $f((1,2))=f((2,4))$.

As numbers, $\frac12$ and $\frac24$ are precisely the same number. Your function $\operatorname{numerator}(x)$ as you use it is not well-defined since it depends on the particular representation of the (rational) number $x$.

The law you're quoting holds only when $f$ is a function.

What you're calling "$\mathit{numerator}(\cdots)$" is not a function, because it depends on something external to which number its input is -- namely on how you've chosen to represent that number.

(Arguably the law is not really a deep thing, but rather part of the definition of what it means to be a function).

Axiomatically, you can define the rational numbers as a field of characteristic 0 which has no proper subfields. You can prove that any two such fields are uniquely isomorphic (so it doesn't matter which of them you choose to call $\mathbb Q$), and the equivalence-classes-of-pairs construction from textbooks shows you that believing in set theory is sufficient to know that there are fields with this property.

• So what is rational number? – Trismegistos Oct 9 '15 at 13:05
• @Trismegistos: Platonically? I'd say it's a real number with the property that you can make an integer by adding it to itself an appropriate number of times. – Henning Makholm Oct 9 '15 at 13:14
• No, I mean formally but I already see you provide answer. This is algebraic definition and I counter on seeing some other definition similar to Peano Axioms. – Trismegistos Oct 9 '15 at 17:47